Computability and Logic

25.2. OPERATIONS IN NONSTANDARD MODELS 311
Let Y + be {n: M |= α*(n)}, and let Y − be its complement, {n: M |= ∼α*(n)}. Then
we have
Y + ={n: for some a, b + = a ⊕···⊕a (π(n) as)}.
If the function ⊕ is recursive, then (much as in the proof of Theorem 25.4a) since
the function g taking a to a ⊕···⊕a (π(n) as) is obtainable from ⊕ by primitive
recursion and composition with π, this g is recursive. Since
Y + ={n: ∃ab + = g(a, n)}.
Y + is semirecursive. A similar argument with b − in place of b + shows that the complement
Y − of Y + is also semirecursive, from which it follows that Y + is recursive.
But this is impossible, since Y + contains A and is disjoint from B.
For the proof of Theorem 25.4c, we need lemmas analogous to those used for
Theorem 25.4a, with ⊗ in place of ⊕. We state these as Lemmas 25.5c through 25.7c
below. These lemmas pertain to exponentiation. Now the notation x y for exponentiation
is not available in L, any more than the notation π for the function enumerating
the primes. But we allow ourselves to use that in stating the lemmas, rather than use
a more correct but more cumbersome formulation in terms of a formula representing
the exponential function. We also write x ↓ y for ‘y has an integral xth root’ or ‘y is
the xth power of some integer’. The only real novelty comes in the proof of Lemma
25.6c, so we give that proof, leaving other details to the reader.
25.5c Lemma. Let M be a nonstandard model of arithmetic. For any m > 0,
M |= ∀xx m = x · ··· · x (m xs).
25.6c Lemma. Let M be a nonstandard model of arithmetic. Let A(x) be any formula
of L. Then there is a nonstandard element d such that
M |= ∃ y∀x < z (∃w((x,w)&w ↓ y) ↔ A(x))[d].
25.7c Theorem. Let M be a nonstandard model of arithmetic. Let A(x) be any formula
of L. Then there exists a b such that for every n,
M |= A(n) if and only if for some a, b = a ⊗···⊗a (π(n) as).
Proof of Lemma 25.6c: It is enough to show that
∀z∃y∀x < z (∃w((x,w)&W ↓ y) ↔ A(x))
is true in N , since it must then be true in M. Recall that we have shown in the proof
of Lemma 25.6b that
∀z∃y∀x < z (∃w((x,w)&w | y) ↔ A(x))
is true in N . It suffices to show, therefore, that the following is true in N :
∀y∃v∀w(w ↓ v ↔ w | y).
In fact, given y, 2 y will do for v (unless y = 0, in which case v = 0 will do). For
suppose w divides y, say y = uw. Then 2 y = 2 uw = (2 u ) w , and 2 y is a wth power.